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We received two particularly good solutions to this problem. The first is from Helen and Daniela who go to Aldermaston C of E Primary School. They drew this table to help answer the questions:
They therefore concluded that:
There are $4$ children behind each chair.
Children $9, 15, 21, 27$ were standing behind chair $3$.
Here's the table they drew for only four chairs:
They comment that:
Seven children are behind $1$ and $2$. Six children are behind $3$ and $4$.
There can't be the same number behind each chair because $30$ doesn't go into $4$ but $30$ goes into $6$. (I think perhaps you mean that $4$ doesn't go into $30$, but $6$ does.)
The second was from Joe and Richard at St. Nicolas CE Junior School, Newbury who also drew a diagram to help them solve the problem:
Joe and Richard say:
We started by drawing children round the circle like in our diagram above. We answered the questions next.
1) Four people were standing behind each chair.
2) Behind number $3$, there were $9, 15, 21, 27$.
We made another drawing but with $4$ chairs and put the children round the circle. In the end, we counted the children up. $1$ and $2$ had $7$ children behind each and $3$ and $4$ had $6$ children. Then we noticed that however many chairs there were, the numbers went up by the highest numbered chair. We tested this solution by using $2$ to $6$ chairs and it worked.
We also noticed that on the last chair, the numbers go up in its table. We tested this on numbers of chairs from $2$ to $6$ and it worked with every single one.
We noticed that because $6$ is a factor of $30$ and $4$ isn't, $4$ didn't have the same number of children around each chair and $6$ did. Then we noticed the only numbers of chairs that would have an equal number of children behind each chair were factors of $30$, which are $1, 2, 3 ,5, 6, 10, 15$ and $30$.
Helen and Pauline from Moorfield Junior School also drew a chart. I think this was a useful strategy. Well done to everyone who answered this correctly and gave good explanations.